A) \[\frac{1}{y}\]
B) \[y\]
C) \[1-y\]
D) \[1+y\]
Correct Answer: B
Solution :
We have, \[\frac{2\sin \alpha }{1+\cos \alpha +\sin \alpha }=y\] Then \[\frac{4\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}{2{{\cos }^{2}}\frac{\alpha }{2}+2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}=y\] Þ \[\frac{2\sin \frac{\alpha }{2}}{\cos \frac{\alpha }{2}+\sin \frac{\alpha }{2}}\times \frac{\left( \sin \frac{\alpha }{2}+\cos \frac{\alpha }{2} \right)}{\left( \sin \frac{\alpha }{2}+\cos \frac{\alpha }{2} \right)}=y\] Þ \[\frac{1-\cos \alpha +\sin \alpha }{1+\sin \alpha }=y\].
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